Question

Find intervals of concavity and points of inflexion for the following functions:

f(x) = x(x – 4)³ 

Answer 1

f'(x) = 3x(x – 4)² + (x – 4)³(1)

= (x – 4)² (4x – 4)

= 4 (x – 4)² (x – 1)

f'(x) = 4 [(x-4)² (1) + (x – 1)² (x – 4)]

= 4(x – 4)(x – 4 + 2x – 2)

= 4(x – 4)(3x – 6)

= 12(x – 4)(x – 2)

f'(x) = 0

⇒ 12 (x – 4) (x – 2) = 0

Critical points x = 2, 4

The intervals are `(- oo, 2)`, (2, 4) and `(4, oo)`

In the interval `(-oo, 2)`, f”(x) > 0 ⇒ Curve is Concave upward

In the interval (2, 4), f”(x) < 0 ⇒ Curve is Concave downward.

In the interval `(4, oo)`, f”(x) > 0 ⇒ Curve is Concave upward.

The curve is concave upward in

`(-oo, 2), (4, oo)` it is concave downward in (2, 4).

f”(x) changes its sign when passing through x = 2 and x = 4

Now f(2) = 2(2 – 4)³

= – 16 and f(4)

= 4(4 – 4)³

= 0

∴ The points of inflection are (2, -16) and (4, 0).